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So far I've done this:

LHS $ =\cos^{2}3x-\sin^{2}3x$

$={(4\cos^{3}x-3\cos{x})}^2 -{(3\sin{x}-4\sin^{3}x)}^2$

$=16\cos^{6}x+9\cos^{2}x-24\cos^{4}x-9\sin^{2}x-16\sin^{6}x+24\sin^{4}x$

I can tell I'm going in the right direction but how should I proceed further?


EDIT I used the identity $\cos{2x}=2\cos^{2}x-1$ to solve it in a simpler way. viz.

LHS $= 2\cos^{2}3x-1$

$=2{(4\cos^{3}x-3\cos{x})}^2-1$

$2(16\cos^{6}x+9\cos^{2}x-24\cos^{4}x)-1$

$=32\cos^{6}x+18\cos^{2}x-48\cos^{4}x-1$

Still thank you for the answers!

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    $\begingroup$ Use $\sin^2 x = 1 - \cos^2 x$. $\endgroup$
    – user49640
    Commented Jun 16, 2017 at 7:05
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    $\begingroup$ use $\sin^2x=1-\cos^2x$ $\endgroup$
    – CY Aries
    Commented Jun 16, 2017 at 7:05
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    $\begingroup$ $\sin^4x=(1-\cos^2x)^2$, etc. $\endgroup$
    – CY Aries
    Commented Jun 16, 2017 at 7:05
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    $\begingroup$ Use $\sin^2x = 1 - \cos^2x$ $\endgroup$
    – Chris
    Commented Jun 16, 2017 at 7:11
  • $\begingroup$ In the general case, the question is related to the Chebishev Polynomials $\endgroup$
    – G Cab
    Commented Jun 16, 2017 at 13:26

5 Answers 5

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Another way:

$$\cos(3\cdot2x)=4\cos^3(2x)-3\cos2x$$

Now use $\cos2x=2\cos^2x-1$

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Note that you can use the identity $\cos^2 x+\sin^2 x=1$(or alternatively, $\sin^2 x=1-\cos^2 x$).

Take each side of the equation to the power of $2,3$ to get $$9 \sin^2 x=9-9\cos^2 x$$ $$24\sin^4 x=24-48 \cos x^2+24 \cos^4 x$$ $$16\sin^6 x=16-48 \cos^2 x+48 \cos^4 x-16\cos^6 x$$ Tedious, but it probably will work.

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  • $\begingroup$ @Arthur Yes, I miswrote that actually. $\endgroup$
    – S.C.B.
    Commented Jun 16, 2017 at 7:25
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Simpler way!

Use the fact $\displaystyle (\cos(x)+i\sin(x))^6 = \cos(6x)+i\sin(6x)$.

Expand, use $\sin^2(x)=1-\cos^2(x)$ to simplify higher powers of sine in the real terms, and then organizing by real terms, we get $\cos(6x)+i\sin(6x)=$

$=32\cos^6(x)-48\cos^4(x)+18\cos^2(x)+32i\sin(x)\cos(x)^5-32i\sin(x)\cos(x)^3+6i\sin(x)\cos(x)-1$

Consider the real part.

This gives you $\cos(6x)=32\cos^6(x)-48\cos^4(x)+18\cos^2(x)-1$.

If you consider the imaginary part, you can isolate $i\sin(6x)$.

This also gives you $\sin(6x)=2\sin(x)\cos(x)^5-32\sin(x)\cos(x)^3+6\sin(x)\cos(x)$

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  • $\begingroup$ @Arthur I think it's okay to point out a quicker way of doing the first part. $\endgroup$
    – user49640
    Commented Jun 16, 2017 at 7:16
  • $\begingroup$ @user49640 Then do that in a comment, not an answer. This question is not asking about how to solve $\cos(6x)$, but how to continue the calculations he's already done. Funny when the answers should've been comments, and the comments contain the answer the OP is looking for. $\endgroup$
    – Arthur
    Commented Jun 16, 2017 at 7:18
  • $\begingroup$ I don't understand this. $\endgroup$
    – Raknos13
    Commented Jun 16, 2017 at 7:18
  • $\begingroup$ @mettledmike This uses De Moivre's formula about complex numbers. $\endgroup$
    – user49640
    Commented Jun 16, 2017 at 7:19
  • $\begingroup$ This is because $e^{ix}=\cos(x)+i\sin(x)$, so that implies that $(e^{ix})^6=e^{6ix}=\cos(6x)+i\sin(6x)=(\cos(x)+i\sin(x))^6$ $\endgroup$ Commented Jun 16, 2017 at 7:20
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Uain Prosthaphaeresis Formulas, $$\cos6x+\cos 2x=2\cos4x\cos2x\iff\cos6x=\cos2x(2\cos4x-1)$$

$\cos4x=2\cos^22x-1$ and $\cos2x=2\cos^2x-1$

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Expand $(\cos(x)+i\sin(x))^6=\cos(6x)+i\sin(6x) $ and take raeal part of both sides

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    $\begingroup$ If you do that, you get the exact same problem that the OP has already got, namely a lot of $\sin^2$ terms that needs to be taken care of, so this is not helpful. $\endgroup$
    – Arthur
    Commented Jun 16, 2017 at 7:13

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